Uzytkownik "Richard Harrison" napisal w
wiadomosci ...
Sz. Bialek wrote:
"Are the radio waves different than light?"

Yes, as light waves are much higher in frequency than radio waves but,
in most ways they are identical. As an example, cross-polarized
receptors for both light and radio waves suffer greatly in receptivity.

So in most ways yes.
Radio waves from the dipole are polarized. Does it mean that light is
emitted by a dipoles?

We can shield the one end of the dipole. Are such waves polarized?

Why the dipoles exhibit the directional pattern?

Are the ansfers in "Electronics and Radio Engineering"?
S*

Pat wrote:
"To the OP, heat is not electromagnetic radiation."

Physics books disagree with Pat. Henry Semat, Ph.D. wrote on page 327 of
"Fundamentals of Physics":
"The transfer of heat by the process of radiation need not involve the
use of material media. An outstanding example is radiation of energy
from the sun to the earth: by far the greatest part of space between
these two bodies is a very good vacuum. The fadiant energy consists of
electromagnetic waves which travel with the speed of light, about
186,000 miles per second."

Best regards, Richard Harrison, KB5WZI

On Mon, 3 May 2010 10:30:26 -0500, (Richard
Harrison) wrote:

"To the OP, heat is not electromagnetic radiation."

Physics books disagree with Pat.

Hi Richard,

The discussion of heat is more a metaphysical concept because it is
confused by our senses.

Entropy demands that everything inexorably cools by dissipating its
energy (heat) into the void of cold space. Hence, everything radiates
(and yet we spend very little time writing about it, except for Art).
A good deal of this entropic radiation is like waiting an infinite
time for a circuit with infinite-1 Q to stop ringing.

The sun burns bright in the cosmos, but the greater part of the cosmos
is unheated by the sun even though all of the cosmos is illuminated
(radiated) by the sun.

Direct observation 1: Put two plates out in the noon sun but
undisturbed by the motion of air. One plate of metal, the other of
glass. Which gets hotter? Same amount of radiation from a known heat
source, but clearly different results in heat.

Direct observation 2: Shine (radiate) an IR LED onto the back of your
hand, or onto your forehead (an excellent heat detector). What do you
feel? At some IR wavelengths, your head is as translucent as an
architectural glass brick.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

Hi Richard,

The discussion of heat is more a metaphysical concept because it is
confused by our senses.

Entropy demands that everything inexorably cools by dissipating its
energy (heat) into the void of cold space. Hence, everything radiates
(and yet we spend very little time writing about it, except for Art).
A good deal of this entropic radiation is like waiting an infinite
time for a circuit with infinite-1 Q to stop ringing.

The sun burns bright in the cosmos, but the greater part of the cosmos
is unheated by the sun even though all of the cosmos is illuminated
(radiated) by the sun.

Direct observation 1: Put two plates out in the noon sun but
undisturbed by the motion of air. One plate of metal, the other of
glass. Which gets hotter? Same amount of radiation from a known heat
source, but clearly different results in heat. . .

This illustrates a classical confusion between heat and temperature,
probably aggravated by our use of "hot" as a description of temperature
rather than heat. Heat is energy. Absorption or transfer of heat results
in a change in temperature. "Hot" (high temperature) objects radiate
more heat than cold objects. The more heat an object, such as a plate,
absorbs, the higher its temperature. Once this basic distinction is
clear, a lot of the mystery disappears.

There are, of course, other mechanisms of heat transfer other than
radiation, namely convection and conduction. But heat transfer has the
same effect on temperature regardless of the mechanism.

When doing experiments with the sun's rays, you sometimes get
non-intuitive results, because there's a lot of energy (heat) at
wavelengths we can't see, particularly at the ultraviolet end. The
reflective or absorptive properties of an object aren't necessarily the
same at infrared or ultraviolet wavelengths as they are at visible
wavelengths. For an example, you can't see the difference in my skin
when coated with sun block or not. But there's sure a difference in
energy (heat) absorption!

Roy Lewallen, W7EL

Roy Lewallen wrote:
ut clearly different results in heat. . .

This illustrates a classical confusion between heat and temperature,
probably aggravated by our use of "hot" as a description of temperature
rather than heat. Heat is energy. Absorption or transfer of heat results
in a change in temperature. "Hot" (high temperature) objects radiate
more heat than cold objects. The more heat an object, such as a plate,
absorbs, the higher its temperature. Once this basic distinction is
clear, a lot of the mystery disappears.

There are, of course, other mechanisms of heat transfer other than
radiation, namely convection and conduction. But heat transfer has the
same effect on temperature regardless of the mechanism.

When doing experiments with the sun's rays, you sometimes get
non-intuitive results, because there's a lot of energy (heat) at
wavelengths we can't see, particularly at the ultraviolet end. The
reflective or absorptive properties of an object aren't necessarily the
same at infrared or ultraviolet wavelengths as they are at visible
wavelengths. For an example, you can't see the difference in my skin
when coated with sun block or not. But there's sure a difference in
energy (heat) absorption!

Roy Lewallen, W7EL


There are also complications about temperature when referring to solids,
liquids, and gases. The "temperature" of even a weekly ionized plasma
is quite high (e.g. 11000 K per eV), but that more to do with the
velocity of the ions and the mean free path. There's not much mass
there, so the "heat" is small. That is, even though the ionosphere is
"hot" in a temperature sense, it's not very "hot" in a sensible transfer
of heat sense.


BTW, I think the sunburn is not from thermal absorption, but from
photons with enough energy to make the reaction go. The total energy
in the UV of sunlight is MUCH lower than the total energy in the visible
range. The power spectrum of sunlight is pretty close to the spectral
sensitivity of your eyes (which evolved that way to match, I would think).

At least one reference says that sunburn is a direct reaction to DNA
damage from UV photons. Melanin protects because it absorbs the UV and
turns it into heat.

http://www.scientificamerican.com/ar...s-when-you-get

Jim Lux wrote:

There are also complications about temperature when referring to solids,
liquids, and gases. The "temperature" of even a weekly ionized plasma
is quite high (e.g. 11000 K per eV), but that more to do with the
velocity of the ions and the mean free path. There's not much mass
there, so the "heat" is small. That is, even though the ionosphere is
"hot" in a temperature sense, it's not very "hot" in a sensible transfer
of heat sense.


BTW, I think the sunburn is not from thermal absorption, but from
photons with enough energy to make the reaction go. The total energy
in the UV of sunlight is MUCH lower than the total energy in the visible
range. The power spectrum of sunlight is pretty close to the spectral
sensitivity of your eyes (which evolved that way to match, I would think).

A good graph of sunlight power density vs wavelength can be found at
http://en.wikipedia.org/wiki/File:Solar_Spectrum.png. Comparing areas of
various graph sections shows that the UV part of the spectrum contains
maybe 1/5 the amount of energy as the visible part -- plenty enough to
embrittle plastics and fabrics and sunburn skin. But the infrared energy
-- invisible to our eyes -- looks to be at least equal to the visible
energy.

At least one reference says that sunburn is a direct reaction to DNA
damage from UV photons. Melanin protects because it absorbs the UV and
turns it into heat.

http://www.scientificamerican.com/ar...s-when-you-get


Some people like to view electromagnetic waves as photons. I find waves
easier to understand, but each to his own.

My explanation was simplified. There's also latent heat or heat of
change of state. For example, if you apply heat to ice, it'll warm up to
0C, but stay at that temperature in spite of the heat input until it
melts. The heat (energy) goes into converting the ice to water instead
of raising the temperature. After it all melts, continued heat
application will of course raise the temperature of the water.(*) Until
it reaches the boiling point, that is. Then the same thing happens again
-- it stays at 100C until it all boils. If you confine the resulting
steam, adding heat will raise both its temperature and pressure after
the water is all converted.

(*) That's why people experienced in cold weather outdoor activities
never eat unmelted snow for water when there's any danger of hypothermia
-- it takes about twice as much heat just to melt 0C snow into 0C water
as it does to raise the temperature of 0C water to body temperature. In
other words, you use up 3 times the energy eating 0C snow as you do
drinking 0C water.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Jim Lux wrote:

There are also complications about temperature when referring to
solids, liquids, and gases. The "temperature" of even a weekly
ionized plasma is quite high (e.g. 11000 K per eV), but that more to
do with the velocity of the ions and the mean free path. There's not
much mass there, so the "heat" is small. That is, even though the
ionosphere is "hot" in a temperature sense, it's not very "hot" in a
sensible transfer of heat sense.


BTW, I think the sunburn is not from thermal absorption, but from
photons with enough energy to make the reaction go. The total energy
in the UV of sunlight is MUCH lower than the total energy in the
visible range. The power spectrum of sunlight is pretty close to the
spectral sensitivity of your eyes (which evolved that way to match, I
would think).

A good graph of sunlight power density vs wavelength can be found at
http://en.wikipedia.org/wiki/File:Solar_Spectrum.png. Comparing areas of
various graph sections shows that the UV part of the spectrum contains
maybe 1/5 the amount of energy as the visible part -- plenty enough to
embrittle plastics and fabrics and sunburn skin. But the infrared energy
-- invisible to our eyes -- looks to be at least equal to the visible
energy.



The plastics degradation is definitely an "athermal" effect (because
adding carbon black to the plastic inhibits it, but doesn't change the
absorbed power very much.

But..
note that the scale is in wavelength and the energy is "per nm" (because
that's how spectrophotometers work). the photons have less energy at
lower wavelength. (or, you could plot it in frequency, and then look at
the watts/Hz to integrate)

If you look at power spectral density (e.g. watts/hz) it actually peaks
up around 1000 nm (near IR). The Wien displacement law says that 5250K
peaks up at about 550 nm, but the power spectral density at 550nm (545
THz) is about 2/3 that at the peak.
By the time you get to 350nm (857 THz), the energy per hz is about 10%
of what it is at the peak (at 950nm)

Running a quick numerical integration... (multiplying the power spectral
density every 50 nm by the frequency range).. I get 0.166 for all
wavelengths shorter than 320nm, 2.09 for 320-670, and 3.6 for 670-4000 nm
(there's a missing integration constant, so the numbers have some scale
factor, but the relative amounts should match..)
for the band around 400nm, I get .26 and for the band at 550 about 0.34
and for around 650 about .32... Yes, it peaks at 550 nm as expected.

On Mon, 03 May 2010 18:28:25 -0700, Jim Lux
wrote:

Roy Lewallen wrote:
Jim Lux wrote:

Hi Jim,

Much of what you write looks like stream-of-consciousness writing.
Did/do you have a point?

The plastics degradation is definitely an "athermal" effect (because
adding carbon black to the plastic inhibits it, but doesn't change the
absorbed power very much.

UV radiation has migrated awary from electron/atom issues to
molecular/ionic bond issues. Calling it "athermal" seems to be
returning the discussion to the metaphysical.

But..
note that the scale is in wavelength and the energy is "per nm" (because
that's how spectrophotometers work). the photons have less energy at
lower wavelength. (or, you could plot it in frequency, and then look at
the watts/Hz to integrate)

What is the significance of changing from wavelength to frequency?
(But?)

If you look at power spectral density (e.g. watts/hz) it actually peaks
up around 1000 nm (near IR). The Wien displacement law says that 5250K
peaks up at about 550 nm, but the power spectral density at 550nm (545
THz) is about 2/3 that at the peak.
By the time you get to 350nm (857 THz), the energy per hz is about 10%
of what it is at the peak (at 950nm)

Running a quick numerical integration... (multiplying the power spectral
density every 50 nm by the frequency range).. I get 0.166 for all
wavelengths shorter than 320nm, 2.09 for 320-670, and 3.6 for 670-4000 nm
(there's a missing integration constant, so the numbers have some scale
factor, but the relative amounts should match..)
for the band around 400nm, I get .26 and for the band at 550 about 0.34
and for around 650 about .32... Yes, it peaks at 550 nm as expected.

Without going into the math, it seems like you disputed a figure you
then discover "as expected." What was the dispute? What wasn't
expected and then came as a surprise?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Mon, 03 May 2010 18:28:25 -0700, Jim Lux
wrote:

Roy Lewallen wrote:
Jim Lux wrote:

Hi Jim,

Much of what you write looks like stream-of-consciousness writing.
Did/do you have a point?

The plastics degradation is definitely an "athermal" effect (because
adding carbon black to the plastic inhibits it, but doesn't change the
absorbed power very much.

UV radiation has migrated awary from electron/atom issues to
molecular/ionic bond issues. Calling it "athermal" seems to be
returning the discussion to the metaphysical.

All in a thread about temperature and heat..
That was actually in response to Roy's original comment
"When doing experiments with the sun's rays, you sometimes get
non-intuitive results, because there's a lot of energy (heat) at
wavelengths we can't see, particularly at the ultraviolet end."
and my response that there actually isn't much energy in the UV end.

Roy commented about sunburn, and I pointed out that the mechanism in
sunburn isn't thermal (and this is important to folks who worry about
RF exposure limits and regulatory compliance.. thermal effects have one
biological result, athermal effects are another..)

My comment was that sunburn (and Roy's example of plastic degradation)
are due to the energy of UV photons actually causing a chemical
reaction, as opposed to making something happen because of heat.





But..
note that the scale is in wavelength and the energy is "per nm" (because
that's how spectrophotometers work). the photons have less energy at
lower wavelength. (or, you could plot it in frequency, and then look at
the watts/Hz to integrate)

What is the significance of changing from wavelength to frequency?
(But?)

Roy's comment was about the amount of energy in the non-visible bands
(presumably in response to my comment that human eye sensitivity tends
to match that of the solar spectrum/ 5250K blackbody), and he cited the
very commonly seen graph in W/nm, with a scale linear in nm.

My point is that in the RF world, we tend look at power spectral density
in terms of W/Hz, so when you are looking at the graphs (with a linear
scale of wavelength or frequency, as apppropriate), a visual estimate of
the "integrated area under the curve" can be misleading. If you plot
the same data, but in W/Hz, and with a scale linear in frequency, you
get a very different looking graph.

Try it.. the equation is of the general form
power density (per hertz) =
constant1*frequency^3/(exp(constant2*frequency/T)-1)

power density (per unit wavelength) =
constant1/lambda^5 * 1/(exp(constant2/(lambda*T))-1)

Thanks, Jim for the correction. I had failed to notice that the graph
scale was normalized to wavelength. I stand corrected.

Roy Lewallen, W7EL